The eigenvalue for a class of singular p-Laplacian

Applied Mathematics and Computation 235 (2014) 412–422

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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

The eigenvalue for a class of singular p-Laplacian fractional differential equations involving the Riemann–Stieltjes integral boundary condition q Xinguang Zhang a,d,⇑, Lishan Liu b, Benchawan Wiwatanapataphee c, Yonghong Wu d a

School of Mathematical and Informational Sciences, Yantai University, Yantai 264005, Shandong, China School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, China c Faculty of Science, Mahidol University, Bangkok 10400, Thailand d Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia b

a r t i c l e

i n f o

a b s t r a c t In this paper, we are concerned with the eigenvalue problem of a class of singular p-Laplacian fractional differential equations involving the Riemann–Stieltjes integral boundary condition. The conditions for the existence of at least one positive solution is established together with the estimates of the lower and upper bounds of the solution at any instant of time. Our results are derived based on the method of upper and lower solutions and the Schauder fixed point theorem. Ó 2014 Elsevier Inc. All rights reserved.

Keywords: Upper and lower solutions p-Laplacian operator Fractional differential equation Integral boundary condition Eigenvalue

1. Introduction This paper deals with the eigenvalue problem for a class of singular p-Laplacian fractional differential equations (PFDE for short) involving the Riemann–Stieltjes integral boundary condition

8 < Db u ðDt a xÞ ðtÞ ¼ kf ðt; xðtÞÞ; t p :

xð0Þ ¼ 0;

a

Dt xð0Þ ¼ 0;

t 2 ð0; 1Þ; R1 xð1Þ ¼ 0 xðsÞdAðsÞ;

ð1:1Þ

where Dbt and Dat are the standard Riemann–Liouville derivatives with 1 < a 6 2; 0 < b 6 1. A is a function of bounded varR1 iation and 0 xðsÞdAðsÞ denotes the Riemann–Stieltjes integral of x with respect to A, the p-Laplacian operator up is defined as

up ðsÞ ¼ jsjp2 s; p > 1, f ðt; xÞ : ð0; 1Þ ð0; þ1Þ ! ½0; þ1Þ is continuous and may be singular at t ¼ 0; 1 and x ¼ 0. Integral and derivative operators of fractional order can describe the characteristics exhibited in many complex processes and systems having long-memory in time, and for this reason many classical integer-order models for complex systems are being substituted by fractional order models. Fractional calculus also provides an excellent tool to describe the hereditary properties of materials and processes, particularly in viscoelasticity, electrochemistry and porous media (see [1–5]). Many successful new applications of fractional calculus in various fields have also been reported recently. For example, Nieto and Pimentel [18] extended a second-order thermostat model to the fractional model, Ding and Jiang [19] used waveform q The authors were supported financially by the National Natural Science Foundation of China (11371221) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017). ⇑ Corresponding author at: School of Mathematical and Informational Sciences, Yantai University, Yantai 264005, Shandong, China. E-mail address: [email protected] (X. Zhang).

http://dx.doi.org/10.1016/j.amc.2014.02.062 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.

X. Zhang et al. / Applied Mathematics and Computation 235 (2014) 412–422

413

relaxation methods to study some fractional functional differential equations models. For the basic theories of fractional calculus and some recent work in application, the reader is referred to Refs. [17,23–34]. Much of the work on fractional calculus deals with boundary value problems [6–9,20]. In particular, Ahmad and Nieto [20] considered a nonlinear Dirichlet boundary value problem of sequential fractional integro-differential equations in the sense of the Caputo fractional derivative, and the existence results are established by means of some standard tools of fixed point theory. On the other hand, some developments on the topic involving the p-Laplacian operator and complex boundary value conditions have been reported [10–13,21]. In [22], Li and Lin considered a Hadamard fractional boundary value problem with a p-Laplacian operator as below:

(

Dbt up ðDt a xÞ ðtÞ ¼ f ðt; xðtÞÞ; xð1Þ ¼ x0 ð1Þ ¼ x0 ðeÞ ¼ 0;

1 < t < e;

a

Dt xð1Þ ¼ Dat xðeÞ ¼ 0;

where 2 < a 6 3; 1 < b 2; up ðsÞ ¼ jsjp2 s; p > 1, and f : ½1; e ½0; þ1Þ ! ½0; þ1Þ is a positive continuous function. By using the Leray–Schauder type alternative and the Guo–Krasnoselskii fixed point theorem, the existence and the uniqueness of the positive solutions were established. The aim of this paper is to deals with the eigenvalue problem for the PFDE involving the Riemann–Stieltjes integral boundary condition, which allows the boundary conditions to be quite general, by using the upper and lower solutions and the Schauder fixed point theorem, so as to determine the interval of eigenvalue for the existence of positive solutions. 2. Preliminaries and lemmas Denote by C½0; 1 the space of all continuous functions on [0, 1] with the usual norm jjxjj ¼ max06t61 jxðtÞj. Indeed, C½0; 1 is a Banach space with a partial order, namely for x; y 2 C½0; 1; x 6 y () xðtÞ 6 yðtÞ, for t 2 ½0; 1. Now consider the linear fractional differential equation

(

Dat xðtÞ þ hðtÞ ¼ 0; xð0Þ ¼ 0;

xð1Þ ¼

t 2 ð0; 1Þ; R1 xðsÞdAðsÞ: 0

ð2:1Þ

Lemma 2.1 [14]. Given hðtÞ 2 L1 ½0; 1, the problem

Dat xðtÞ þ hðtÞ ¼ 0; xð0Þ ¼ 0;

0 < t < 1;

ð2:2Þ

xð1Þ ¼ 0

has the unique solution

xðtÞ ¼

Z

1

Gðt; sÞhðsÞds;

ð2:3Þ

0

where Gðt; sÞ is the Green function of (2.2) and is given by

1 Gðt; sÞ ¼ CðaÞ

(

½tð1 sÞa1 ;

0 6 t 6 s 6 1;

½tð1 sÞa1 ðt sÞa1 ; 0 6 s 6 t 6 1:

ð2:4Þ

Consider the problem

Dat xðtÞ ¼ 0; xð0Þ ¼ 0;

0 < t < 1;

ð2:5Þ

xð1Þ ¼ 1;

by the property of the Riemann–Liouville fractional integral and derivative operators, the unique solution of the problem R1 (2.5) is ta1 . Defining G A ðsÞ ¼ 0 Gðt; sÞdAðtÞ, as in [15,16], we can get that the Green function for the nonlocal FDE (2.1) is given by

Jðt; sÞ ¼

t a1 G A ðsÞ þ Gðt; sÞ; 1C

C¼

Z

1

t a1 dAðtÞ:

ð2:6Þ

0

Lemma 2.2 [16]. Let 0 6 C < 1 and G A ðsÞ 0 for s 2 ½0; 1, then the Green function defined by (2.6) satisfies: (1) Jðt; sÞ > 0; for all t; s 2 ð0; 1Þ. (2) There exist two constants c ; c such that

c t a1 G A ðsÞ 6 Jðt; sÞ 6 c ta1 6 c ;

t; s 2 ½0; 1;

ð2:7Þ

414


where

c ¼

1 ; 1C

jjG A ðsÞjj 1 þ : 1C Cða 1Þ

c ¼

Let q > 1 satisfy the relation 1q þ 1p ¼ 1, where p is given by (1.1). To study the PFDE (1.1), we first consider the associated linear PFDE

8 < Db u ðDt a xÞ ðtÞ þ hðtÞ ¼ 0; t p :

Dat xð0Þ ¼ 0;

xð0Þ ¼ 0;

t 2 ð0; 1Þ; R1 xð1Þ ¼ 0 xðsÞdAðsÞ

ð2:8Þ

for h 2 L1 ½0; 1 and h 0. For convenience, let b ¼ CðbÞ1 , then we have the following lemma. Lemma 2.3. The associated linear PFDE (2.8) has the unique positive solution

xðtÞ ¼

Z

1

Jðt; sÞ

Z

0

s

bðs sÞb1 hðsÞds

ds:

ð2:9Þ

0

Proof. In fact, let w ¼ Dat x;

(

q1

Dbt v ðtÞ þ hðtÞ ¼ 0; v ð0Þ ¼ 0

v ¼ up ðwÞ. Then the solution of the initial value problem t 2 ð0; 1Þ;

ð2:10Þ

is given by v ðtÞ ¼ C 1 t b1 Ib hðtÞ; t 2 ½0; 1. From the relations v ð0Þ ¼ 0; 0 < b 6 1, we get C 1 ¼ 0, and consequently

v ðtÞ ¼ Ib hðtÞ;

t 2 ½0; 1:

ð2:11Þ

a

Noting that Dt x ¼ w; w ¼ u1 p ðv Þ, we have from (2.11) that the solution of (2.8) satisfies

(

b Dat xðtÞ ¼ u1 t 2 ð0; 1Þ; p ðI hðtÞÞ; R1 xð0Þ ¼ 0; xð1Þ ¼ 0 xðsÞdAðsÞ:

ð2:12Þ

By (2.6), the solution of Eq. (2.12) can be written as

xðtÞ ¼

Z

1 b Jðt; sÞu1 p ðI hðsÞÞds;

0

t 2 ½0; 1:

b q1 b Since hðsÞ P 0; s 2 ½0; 1, we have u1 ; s 2 ½0; 1, which implies that the solution of Eq. (2.8) is p ðI hðsÞÞ ¼ I hðsÞ

xðtÞ ¼

Z

1

Jðt; sÞ

Z

0

s

bðs sÞb1 hðsÞds

q1

ds;

t 2 ½0; 1:

0

Definition 2.3. A continuous function WðtÞ is called a lower solution of the PFDE (1.1) if it satisfies

8 < Db u ðDt a WÞ ðtÞ 6 kf ðt; WðtÞÞ; t p :

t 2 ð0; 1Þ; R1 Wð1Þ P 0 WðsÞdAðsÞ:

a

Wð0Þ P 0;

Dt Wð0Þ P 0;

Definition 2.4. A continuous function UðtÞ is called a upper solution of the PFDE (1.1) if it satisfies

8 < Db u ðDt a UÞ ðtÞ P kf ðt; UðtÞÞ; t p :

Uð0Þ 6 0;

a

Dt Uð0Þ 6 0;

t 2 ð0; 1Þ; R1 Uð1Þ 6 0 UðsÞdAðsÞ:

Remark 2.1. Assume 0 6 C < 1 and G A ðsÞ 0 for s 2 ½0; 1, and x 2 Cð½0; 1; RÞ satisfies

xð0Þ ¼ 0;

xð1Þ ¼

Z

1

xðsÞdAðsÞ

0

and Dat xðtÞ 0 for any t 2 ½0; 1. Then xðtÞ 0; t 2 ½0; 1. In fact, let

Dat xðtÞ ¼ hðtÞ

ð2:13Þ

415


and

xð0Þ ¼ 0;

xð1Þ ¼

Z

1

xðsÞdAðsÞ:

ð2:14Þ

0

Then the unique solution of Eq. (2.13) subject to boundary condition (2.14) is

xðtÞ ¼

Z

1

Jðt; sÞhðsÞds:

0

By Lemma 2.2 (1) and noticing hðtÞ 0, one gets the conclusion xðtÞ 0; t 2 ½0; 1. Lemma 2.4 (Schauder fixed point theorem). Let T be a continuous and compact mapping of a Banach space E into itself, such that the set

fx 2 E : x ¼ rTx; for some 0 6 r 1g

ð2:15Þ

is bounded. Then T has a fixed point.

3. Main results To establish the existence of a solution to the boundary value problem (1.1), we need to make the following assumptions. (H0) A is a function of bounded variation satisfying G A ðsÞ 0 for s 2 ½0; 1 and 0 6 C < 1. (H1) f : ð0; 1Þ ð0; þ1Þ ! ½0; þ1Þ is continuous and is non-increasing in x > 0; and for all r 2 ð0; 1Þ, there exists a constant > 0 such that, for any ðt; xÞ 2 ð0; 1 ð0; þ1Þ,

f ðt; rxÞ 6 r f ðt; xÞ:

ð3:1Þ

Remark 3.1. For all r > 1, we have

1 r

2 ð0; 1Þ and y ¼ rx 2 ð0; 1Þ, given x 2 ð0; 1Þ, and thus from (H1), we have

1 1 f t; y 6 f ðt; yÞ r r using y ¼ rx, we have

f ðt; xÞ 6

1 f ðt; rxÞ; r

which give

f ðt; rxÞ P r f ðt; xÞ:

ð3:2Þ

Now denote

gðsÞ ¼ f ðs; sa1 Þ;

for any s 2 ð0; 1Þ 1

and introduce the Lebesgue space Ll ½0; 1 ð0 < l < 1Þ with the norm denoted by

jjgjj

1

Ll

¼

Z

1

l 1 g l ðsÞds :

0

Theorem 3.1. Suppose (H0) and (H1) hold, and there exists a constant 0 < l < b such that 1

g 2 Ll ½0; 1:

ð3:3Þ

Then there exists a constant k > 0 such that the PFDE (1.1) has at least one positive solution wðtÞ for any k 2 ðk ; þ1Þ, and morea1 over there exist two constants 0 < l < 1 and L > 1 such that lt 6 wðtÞ 6 Lta1 : Proof. Let E ¼ C½0; 1, and define a subset P of E as follows:

n o a1 1 P ¼ x 2 E : there exists a constant 0 < l < 1such that lt 6 xðtÞ 6 l ta1 ; t 2 ½0; 1 : Clearly, P is a nonempty set since t a1 2 P. Now define the operator T k in E:

416


T k xðtÞ ¼ k

Z

1

Jðt; sÞ

Z

0

q1 bðs sÞb1 f ðs; xðsÞÞds ds;

s

t 2 ½0; 1:

ð3:4Þ

0

We assert that T k is well defined and T k ðPÞ P. In fact, for any x 2 P, there exists a positive number 0 < lx < 1 such that 1 lx t a1 6 xðtÞ 6 lx ta1 ; t 2 ½0; 1, and thus, by Lemma 2.2, ðH1Þ, (3.1) and (3.3) and the Hölder inequality and noticing 0 < l < b, one gets

T k xðtÞ ¼ k

Z

Z

1

Jðt; sÞ

0 1

Jðt; sÞ

6k

Z

bðs sÞb1 f ðs; xðsÞÞds s

bðs sÞb1 f ðs; lx sa1 Þds

0

ðq1Þ q1 a1

b

6 klx 6

ct

Z

Z

1

0

ðq1Þ q1 a1 klx b ct

ðq1Þ q1 a1

b

6 klx

ct

b

ct

ds

q1 ds

s

ðs sÞb1 gðsÞds

q1 ds

0

Z 1 "Z 0

s

ðs sÞ

b1 1l

ds

1l Z

0

jjgjjq1 1

Z

Ll

ðq1Þ q1 a1

6 klx

q1

0

Z

0

s

Z

Ll

1

g l ðsÞds

l #q1

ds

ð3:5Þ

0 1

0

jjgjjq1 1

s

1

Z

0

s

b1

ðs sÞ1l ds

0 bl 1 l 1 s l bl

ð1lÞðq1Þ

ð1lÞðq1Þ

ds

ds

ðq1Þð1lÞ ðq1Þ q1 a1 1 l 6 klx b ct jjgjjq1 < þ1: 1 bl Ll On the other hand, from (3.2) and (2.7), we have

ðT k xÞðtÞ ¼ k

Z Z

1

0

Z s q1 Jðt; sÞ bðs sÞb1 f ðs; xðsÞÞds ds 0

Z s q1 1 Jðt; sÞ bðs sÞb1 f ðs; lx sa1 Þds ds

1

Pk 0

0

ðq1Þ

P klx

b

q1

c ta1

Z

1

G A ðsÞ

Z

0

s

ðs sÞb1 gðsÞds

ð3:6Þ

q1 ds:

0

Choose

~lx ¼ min

8 ( )1 ðq1Þð1lÞ k1 , from (3.5), we have

Z

1

Z s q1 Z Jðt; sÞ bðs sÞb1 f ðs; keðsÞÞds ds 6

0

0

1

Jðt; sÞ

0

Z

Z

s

bðs sÞb1 f ðs; k1 eðsÞÞds

0

1

6

Jðt; sÞ

Z

0

0

0

0

s

bðs sÞb1 f ðs; sa1 Þds

q1 ds

q1 ds

Z s q1 Z 1 Jðt; sÞ bðs sÞb1 gðsÞds ds < þ1 ¼ and

eðtÞ 6 c b

Z

q1

q1

Now let q ¼ c b

1

0

Z

s

ðs sÞb1 gðsÞds

q1

q1

ds 6 c b

0

ðq1Þð1lÞ

1l bl

ðq1Þð1lÞ 1l jjgjjq1 < þ1: 1 bl Ll

jjgjjq1 þ 1 and take 1 l

L 8 " 9 1 Z s q1 #ðq1Þ1 Z 1 < = k ¼ max k1 ; c qðq1Þ G A ðsÞ bðs sÞb1 f ðs; 1Þds ds : : ; 0 0

Then by (2.7) and (3.2), we obtain

þ1 > k

Z

1

Jðt; sÞ

0

P ðk Þ

Z

s

bðs sÞb1 f ðs; k eðsÞÞds

1ðq1Þ

c t a1

Z

1

Z s q1 G A ðsÞ bðs sÞb1 f ðs; eðsÞÞds ds

1ðq1Þ

c t a1

0

Z s q1 1 G A ðsÞ bðs sÞb1 f ðs; qÞds ds

Z 0

P ðk Þ

ds

0

0

P ðk Þ

q1

1ðq1Þ

ð3:10Þ

0

c qðq1Þ t a1

Z

1

G A ðsÞ

0

Z

s

bðs sÞb1 f ðs; 1Þds

q1 ds

0

P t a1 ; 8t 2 ½0; 1: Let

UðtÞ ¼ k eðtÞ;

WðtÞ ¼ k

Z

1

Jðt; sÞ

Z

0

s


q1

ds;

0

then

UðtÞ ¼ T k ðta1 Þ;

WðtÞ ¼ T k ðUðtÞÞ:

ð3:11Þ

It follows from (3.9) and (3.10) that for any t 2 ½0; 1,

UðtÞ ¼ k

Z

1

0

WðtÞ ¼ k

Z

0

Z s q1 Jðt; sÞ bðs sÞb1 gðsÞds ds P k1 eðtÞ P ta1 ; 0

1

Z s q1 Jðt; sÞ bðs sÞb1 f ðs; k eðsÞÞds ds P t a1 :

ð3:12Þ

0

Moreover, by (3.8) and (3.11), we know

Uð0Þ ¼ 0;

Wð0Þ ¼ 0;

Dat Uð0Þ ¼ 0; Dat Wð0Þ ¼ 0;

Uð1Þ ¼

Z

1

UðsÞdAðsÞ;

0

Wð1Þ ¼

Z

ð3:13Þ

1

WðsÞdAðsÞ:

0

Proceeding as in (3.5)–(3.7), we get that UðtÞ; WðtÞ 2 P. By (3.11) and (3.12), we have

ta1 6 WðtÞ ¼ ðT k UÞðtÞ;

t a1 6 UðtÞ;

8t 2 ½0; 1;

ð3:14Þ

418


which implies WðtÞ ¼ ðT k UÞðtÞ ¼ k

Z

1

Jðt; sÞ

Z

0

s


q1

ds 6 k

0

Z

1

Jðt; sÞ 0

Z

s

bðs sÞb1 gðsÞds

q1

ds ¼ UðtÞ; 8t 2 ½0;1:

ð3:15Þ

0

Thus, taking account of f being non-increasing in x > 0, and by (3.11), (3.8), (3.14) and (3.15), we have

Dbt up ðDt a WÞ ðtÞ þ k f ðt; WðtÞÞ ¼ Dbt up ðDt a ðT k UÞÞ ðtÞ þ k f ðt; WðtÞÞ P Dbt up ðDt a ðT k UÞÞ ðtÞ þ k f ðt; UðtÞÞ ¼ k f ðt; UðtÞÞ þ k f ðt; UðtÞÞ ¼ 0;

ð3:16Þ

Dbt up ðDt a UÞ ðtÞ þ k f ðt; UðtÞÞ ¼ Dbt up Dt a T k ðt a1 Þ þ k f ðt; UðtÞÞ ¼ k f ðt; t a1 Þ þ k f ðt; UðtÞÞ P k f ðt; t a1 Þ þ k f ðt; ta1 Þ ¼ 0:

ð3:17Þ

It follows from (3.13) and (3.15)–(3.17) that WðtÞ; UðtÞ are upper and lower solutions of the PFDE (1.1), and WðtÞ; UðtÞ 2 P. Now let us define a function

8 f ðt; WðtÞÞ; x < WðtÞ; > > < Fðt; xÞ ¼ f ðt; xðtÞÞ; WðtÞ 6 x 6 UðtÞ; > > : f ðt; UðtÞÞ; x > UðtÞ:

ð3:18Þ

It then follows from (H1) and (3.18) that F : ð0; 1Þ ½0; þ1Þ ! ½0; þ1Þ is continuous. We now show that the fractional boundary value problem

8 < Db u ðDt a xÞ ðtÞ ¼ k Fðt; xðtÞÞ; t p :

Dat xð0Þ ¼ 0;

xð0Þ ¼ 0;

xð1Þ ¼

t 2 ð0; 1Þ; R1 0

ð3:19Þ

xðsÞdAðsÞ

has a positive solution. Define the operator Dk by

Dk xðtÞ ¼ k

Z

1

Jðt; sÞ

Z

0

s

bðs sÞb1 Fðs; xðsÞÞds

q1

ds;

t 2 ½0; 1:

ð3:20Þ

0

Then Dk : C½0; 1 ! C½0; 1, and a fixed point of the operator Dk is a solution of the PFDE (3.19). On the other hand, from the definition of F and the fact that the function f ðt; xÞ is nonincreasing in x, we obtain f ðt; UðtÞÞ 6 Fðt; xðtÞÞ 6 f ðt; WðtÞÞ provided that WðtÞ 6 x 6 UðtÞ, Fðt; xðtÞÞ ¼ f ðt; WðtÞÞ provided that x < WðtÞ, and Fðt; xðtÞÞ ¼ f ðt; UðtÞÞ provided that x > UðtÞ. So we have

f ðt; UðtÞÞ 6 Fðt; xðtÞÞ 6 f ðt; WðtÞÞ;

8x 2 E:

ð3:21Þ

Furthermore, by (3.14) and (3.21), we have

f ðt; UðtÞÞ 6 Fðt; xðtÞÞ 6 f ðt; t a1 Þ ¼ gðtÞ;

8x 2 E;

ð3:22Þ

It follows from (2.7) and (3.22) that for any x 2 E,

Dk xðtÞ ¼ k

Z

1

Z s q1 Jðt; sÞ bðs sÞb1 Fðs; xðsÞÞds ds

0

6 k c b

0 q1

Z 0

6 k c b

q1

1 Z

s

ðs sÞb1 gðsÞds

q1

ds

ð3:23Þ

0

ðq1Þð1lÞ 1l jjgjjq1 < þ1; 1 bl Ll

namely, the operator Dk is uniformly bounded. On the other hand, let X E be bounded. As the function Jðt; sÞ is uniformly continuous on ½0; 1 ½0; 1, Dk ðXÞ is equicontinuous. By the Arzela–Ascoli theorem, we have Dk : E ! E is completely continuous. Moreover, (3.23) implies that (2.15) holds, thus, by using the Schauder fixed point theorem, Dk has at least one fixed point w such that w ¼ Dk w. Now we prove

WðtÞ 6 wðtÞ 6 UðtÞ;

t 2 ½0; 1:

419


Since w is a fixed point of Dk , by (3.13) and (3.19), we have

wð0Þ ¼ 0;

Uð0Þ ¼ 0;

Dat wð0Þ ¼ 0; Dat Uð0Þ ¼ 0;

wð1Þ ¼

Uð1Þ ¼

Z Z

1

wðsÞdAðsÞ;

0

ð3:24Þ

1

UðsÞdAðsÞ:

0

From (3.8), (3.11), (3.22) and noticing that w is a fixed point of Dk , we also have

Dbt up ðDt a UÞ ðtÞ Dbt up ðDt a wÞ ðtÞ ¼ Dbt up ðDt a UÞ up ðDt a wÞ ðtÞ ¼ k f ðt;t a1 Þ þ k Fðt;wðtÞÞ 6 0; 8t 2 ½0;1:

Let zðtÞ ¼ up ðDt a UðtÞÞ up ðDt a wðtÞÞ. Then

Dbt zðtÞ ¼ Dbt up ðDt a UÞ ðtÞ Dbt up ðDt a wÞ ðtÞ 6 0; t 2 ½0; 1; Dbt zð0Þ ¼ Dbt up ðDt a Uð0ÞÞ Dbt up ðDt a wð0ÞÞ ¼ 0:

It follows from (2.10) and (2.11) that

zðtÞ 6 0 and

up ðDt a UðtÞÞ up ðDt a wðtÞÞ 6 0: Noticing that up is monotone increasing, we have

Dat UðtÞ 6 Dat wðtÞ;

Dat ðU wÞðtÞ 6 0:

i:e:;

It follows from Remark 2.1 and (3.24) that

UðtÞ wðtÞ P 0: Thus we have wðtÞ 6 UðtÞ on ½0; 1. By the same way, we also have wðtÞ P WðtÞ on ½0; 1. So

WðtÞ 6 wðtÞ 6 UðtÞ;

t 2 ½0; 1:

ð3:25Þ

Consequently, Fðt; wðtÞÞ ¼ f ðt; wðtÞÞ; t 2 ½0; 1. Hence wðtÞ is a positive solution of the problem (1.1). Finally, by (3.25) and U; W 2 P, we have 1

lW ta1 6 WðtÞ 6 wðtÞ 6 UðtÞ 6 lU ta1 :

In order to illustrate the validity of our results, we give the following example. Example 3.1. Consider a singular PFDE involving the Riemann–Stieltjes integral boundary condition

! 8 n 3 X 1 > > c 2 2 < Dt up Dt x ðtÞ ¼ k a0 ðtÞ þ ai ðtÞx i ; > > :

t 2 ð0; 1Þ;

i¼1

xð0Þ ¼ 0;

3 2

Dt xð0Þ ¼ 0;

xð1Þ ¼

R1 0

ð3:26Þ

xðsÞdAðsÞ;

where up ðsÞ ¼ jsjp2 s; p > 1; a0 ðtÞ and ai ðtÞ are nonnegative and continuous on ð0; 1Þ; ci P 0 for i ¼ 1; 2; . . . ; n, and

8 1 > < 0; t 2 0; 2; AðtÞ ¼ 2; t 2 12 ; 34 ; >

: 1; t 2 34 ; 1 : Using Theorem 3.1 for the PFDE (3.26), we have the following conclusion. Corollary 3.1. If there exists a constant 0 < l < 12 such that

Z 0

1

" a0 ðtÞ þ

n X ci ai ðtÞt 2 i¼1

#l1 ds < þ1:

ð3:27Þ

420


Then there exists a constant k > 0 such that for any k 2 ðk ; þ1Þ, the PFDE (3.26) has at least one positive solution wðtÞ, and there exist two constants, 0 < l < 1 and L > 1, such that 1

1

lt2 6 wðtÞ 6 Lt2 : Proof. According to the property of the Riemann–Stieltjes integral, the PFDE (3.26) becomes the following four-point boundary value problem with coefficients of both signs

! 8 n 3 X 1 > > c 2 2 < Dt up Dt x ðtÞ ¼ k a0 ðtÞ þ ai ðtÞx i ; > > :

t 2 ð0; 1Þ;

i¼1

xð0Þ ¼ 0;

xð1Þ ¼ 2x 12 x 34 :

3

D2t xð0Þ ¼ 0;

Let a ¼ 32 ; b ¼ 12 ; f ðt; xÞ ¼ a0 ðtÞ þ

Pn

ci , i¼1 ai ðtÞx

then

12 12 1 1 3 C¼ ta1 dAðtÞ ¼ 2 ¼ 0:5482 < 1 2 4 0 Z

and by simple computation, we have G A ðsÞ P 0, and so (H0) holds. On the other hand, let ¼ max16i6n ci , then for all positive numbers r < 1, we have

f ðt; rxÞ ¼ a0 ðtÞ þ

n X ai ðtÞðrxÞci 6 r f ðt; xÞ i¼1 1

and by (3.27), we also have g 2 Ll ½0; 1, where

gðtÞ ¼ a0 ðtÞ þ

n X ci ai ðtÞt 2 : i¼1

So (H1) and (3.3) hold. By Theorem 3.1 there exists a constant k > 0 such that for any k 2 ðk ; þ1Þ, the PFDE (3.26) has at least one positive solution wðtÞ, and there exist two constants, 0 < l < 1 and L > 1, such that 1

1

lt2 6 wðtÞ 6 Lt2 :

Remark 3.2. The formula (3.27) not only implies that ai ðtÞ; i ¼ 0; 1; . . . ; n can be singular on [0, 1], but also f can be singular at x ¼ 0. So far, for the fractional differential equations involving p-Laplacian operator, no results have been obtained in existing work, especially, for the PFDE (1.1) in which f has singularity at t 2 ½0; 1 and x ¼ 0. Remark 3.3. There are a large number of functions that satisfy the conditions of Theorem 3.1. In particular, in Example 3.1, if a0 ðtÞ is nonnegative and continuous on ½0; 1 and ai ðtÞ ¼ t di ; di P c2i , then only f is singular at x ¼ 0. In this case, Corollary 3.1 still holds without the restriction of (3.27). To end this section, we demonstrate the calculation of k by considering the following specific case for Eq. (3.26),

3 8 1 < D2t u3 D2t x ðtÞ ¼ kx16 ðtÞ; :

3 2

xð0Þ ¼ 0; Dt xð0Þ ¼ 0;

Choose 0 < l ¼ 13 < 12 ;

Gðt; sÞ ¼

t 2 ð0; 1Þ; xð1Þ ¼ 2x 12 x 34 :

¼ 16, then (3.27) and (H1) are satisfied, and the corresponding Green function is

8 1 ½tð1sÞ2 > > < G1 ðt; sÞ ¼ Cð3Þ ;

0 6 t 6 s 6 1;

2

> > : G2 ðt; sÞ ¼

1 1 ½tð1sÞ2 ðtsÞ2

Cð32Þ

;

0 6 s 6 t 6 1:

Consequently, we have

8 pffiffiffi pffiffi 1 1 1 8 1 3 > 2 23 ð1 sÞ2 2 12 s 2 þ 34 s 2 ; 0 6 s < 12 ; > 1 > 2G2 2 ; s G2 4 ; s ; 0 6 s < 2 ; > > > > > > < 1 3 1 1 1 < pffiffiffi pffiffi3 1 3 1 G A ðsÞ ¼ 2G1 2 ; s G2 4 ; s ; 2 6 s < 4 ; ¼ 3 2 2 ð1 sÞ2 þ 34 s 2 ; 6 s < 34 ; 2 > > C > > 2 > > : > > pffiffi 2G1 12 ; s G1 34 ; s ; 34 6 s 6 1: 1 > : pffiffiffi 3 2 23 ð1 sÞ2 ; 6 s 6 1: 4 pffiffi 1 1 A ðsÞjj Clearly, 0 6 G A ðsÞ 6 C 23 , and c ¼ 1C ¼ 10:5482 2:2134; c ¼ jjG1C þ C 13 4:6601; b ¼ C 11 1:7725. Noticing that ð2Þ ð2Þ ð2Þ

ð3:28Þ


eðtÞ ¼

Z

1

Jðt; sÞ 0 1

2:2134t 2

Z Z

s

1

0 1

G A ðsÞ

Z

0 1

¼ 2:2134t 2

Z

1

1:7725ðs sÞ2 s12 ds s

12

421

ds

1

1

1:7725ðs sÞ2 s12 ds

12 ds

0 1

5

G A ðsÞs24

0 1

Z

Z

1

1

1

1:7725ð1 sÞ2 s12 ds

0 1

5

1

Z

12

1

ds 5

¼ 2:2134 1:9341t 2 G A ðsÞs24 ds ¼ 4:2809t 2 G A ðsÞs24 ds 0 0 " ! pffiffiffi 12 12 # Z 1 pffiffiffi 2 1 3 1 3 1 5 2 ð1 sÞ 2 s24 ds 2 ¼ 4:8306t 2 s þ s 2 4 2 0 ! 12 # Z 3 " pffiffiffi pffiffiffi! Z 1 pffiffiffi pffiffiffi! 4 1 1 5 3 3 3 5 þ s 2 2 ð1 sÞ2 þ s24 ds þ ð1 sÞ2 s24 ds 1 3 4 2 2 2 4 1

1

¼ 4:8306t 2 ð0:1665 0:3196 þ 0:0748 þ 0:2445 þ 0:0755 þ 0:0441Þ ¼ 1:3806t2 ; 1

we have 1:38061 eðtÞ P t2 , and thus (3.9) is satisfied by taking k1 ¼ 1. Further

Z

1

G A ðsÞ

Z

0

s


q1

12 Z 1 1 1 1 1:7725ðs sÞ2 ds ds ¼ ð2 1:7725Þ2 G A ðsÞs4 ds 0 0 0 12 12 # Z 1 " pffiffiffi pffiffiffi! 2 1 1 1 3 3 1 2 2 ð1 sÞ 2 s4 ds ¼ ð2 1:7725Þ s þ s 2 2 4 2 0 ! 12 # Z 3 " pffiffiffi pffiffiffi! Z 1 pffiffiffi pffiffiffi! 4 1 1 1 3 3 3 1 þ s 2 2 ð1 sÞ2 þ s4 ds þ ð1 sÞ2 s4 ds 1 3 4 2 2 2 4

ds ¼

0

Z

1

G A ðsÞ

Z

s

¼ 1:8828 ð0:2738 þ 0:2286 þ 0:0733 0:2970Þ ¼ 0:5247 and

q ¼ c bq1

!1 Z ðq1Þð1lÞ 19 1 1 13 3 1 1l 14 2 jjgjjq1 þ 1 ¼ 4:6601 1:7725 t dt þ 1 ¼ 11:8394: 1 1 bl 13 Ll 0 2

"

Z

Thus

c qðq1Þ

1

G A ðsÞ

Z

0

s


q1

1 #ðq1Þ1

ds

h i12 1 11 ¼ 2:2134 11:839412 0:5247 1:0634;

0

and hence

k ¼ maxf1; 1:0634g ¼ 1:0634: So for any k > 1:0634, Eq. (3.28) has at lest one positive solution wðtÞ. Moreover by (3.14) and (3.23), we have

ta1 6 WðtÞ 6 wðtÞ 6 UðtÞ 6 k c b

q1

ðq1Þð1lÞ 1l a1 jjgjjq1 ; 1 t bl Ll

consequently, the positive solution wðtÞ satisfies 1

1

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